Change in direction of electric field on conductor surface

[Noaha]

I would like to know why electric field of EM waves changes it's direction to opposite when hitting a conductor surface. I know that electric field inside a conductor is zero but I am not able to connect these two situations.

 

[tech99]

I would like to know why electric field of EM waves changes it's direction to opposite when hitting a conductor surface. I know that electric field inside a conductor is zero but I am not able to connect these two situations.

The usual explanation seems to be that the magnetic field of the incoming wave at the surface is not zero, and induces a surface current. This then creates a surface electric field in the reversed direction.
My own suggestion is that the electrons in the metal are accelerated by the incoming E-field and create a new radiated E-field in the reversed direction.

 

[Noaha]

The usual explanation seems to be that the magnetic field of the incoming wave at the surface is not zero, and induces a surface current. This then creates a surface electric field in the reversed direction.
My own suggestion is that the electrons in the metal are accelerated by the incoming E-field and create a new radiated E-field in the reversed direction.

i didnt understand this part.. " create a new radiated E-field in the reversed direction." Can u explain it a bit more..@tech99

 

[Noaha]

i didnt understand this part.. " create a new radiated E-field in the reversed direction." Can u explain it a bit more..@tech99

In this line "create a new radiated E-field in the reversed direction", did you mean that when an electric field tries to enter the conductor, the electrons in it starts moving in direction opposite to electric field and hence creating a field due to separation of charges in the conductor?

 

[tech99]

In this line "create a new radiated E-field in the reversed direction", did you mean that when an electric field tries to enter the conductor, the electrons in it starts moving in direction opposite to electric field and hence creating a field due to separation of charges in the conductor?

Not charge separation. The free charges on the surface of the metal have an electric field normal to the surface. When the charges are accelerated, these field lines slope back, like a ship's wake.This gives them a component parallel to the surface, which is the new radiated (reflected) field, and it goes the opposite way to the incident, or accelerating, field.
The two cancel out near the surface, so you might think that the charges will not move. But it is like Newton's reaction and re-action; they are in equilibrium as a consequence of the acceleration.

 

[Noaha]

Not charge separation. The free charges on the surface of the metal have an electric field normal to the surface. When the charges are accelerated, these field lines slope back, like a ship's wake.This gives them a component parallel to the surface, which is the new radiated (reflected) field, and it goes the opposite way to the incident, or accelerating, field.
The two cancel out near the surface, so you might think that the charges will not move. But it is like Newton's reaction and re-action; they are in equilibrium as a consequence of the acceleration.

I have one more doubt. From your above answer, did you mean that the incident electric field and the electric field of charges in conductor vectorially add to produce the ship's wack like electric field whose parallel component is the reflected electric field?

 

[tech99]

I have one more doubt. From your above answer, did you mean that the incident electric field and the electric field of charges in conductor vectorially add to produce the ship's wack like electric field whose parallel component is the reflected electric field?

No, the field of the accelerated electron alone resembles a ship's wake. The two fields, incident and reflected, add vectorially, giving zero close to the surface but giving twice the field at certain greater distances.

 

[Vanadium 50]

Do you know method of images? If you do, there's a slick way to see this. (And if you don't, it will just confuse you, which is why I am not writing it)

 

[tech99]

Do you know method of images? If you do, there's a slick way to see this. (And if you don't, it will just confuse you, which is why I am not writing it)

Thank you, but I am sorry I am just an amateur scientist and do not know the method of images.

 

[Vanadium 50]

Yes, but you're not the OP, are you? You're surely not suggesting I not answer his question because you wouldn't understand it?

 

[Noaha]

No, the field of the accelerated electron alone resembles a ship's wake. The two fields, incident and reflected, add vectorially, giving zero close to the surface but giving twice the field at certain greater distances.

I really don't understand..How it can becoming twice at greater distance when it is almost zero on surface?

 

[Noaha]

Do you know method of images? If you do, there's a slick way to see this. (And if you don't, it will just confuse you, which is why I am not writing it)

Can you please explain it?

 

[tech99]

Can you please explain it?

If you just think about a simple mirror. At the surface, the incident and reflected rays are 180 degrees out of phase, so they cancel. At a small height above the surface, the incident ray is slightly phase advanced, and the reflected ray is slightly phase delayed, so they no longer cancel.
At a height of a quarter of a wavelength, the incident ray is advanced in phase by 90 degrees and the reflected ray is delayed in phase by 90 degrees. So the time-of-travel has inserted 180 degree phase shift. Add to this the 180 degrees at the reflection point and you find that they are 360 degrees out of phase. In other words, in phase. And the two electric fields add vectorially, giving twice the field strength.

 

[rumborak]

FYI, there is another way of looking at this, which is the superposition principle (a very useful principle overall if I might add).
So, you know that the electric field inside the conductor is zero. The way to look at it is by rather saying "the electric field *does* permeate the conductor, but additionally, the conductor creates a field inside that is the exact opposite."
The net result (of course) is still that the field inside is zero, but now you can look at the field *outside* the conductor as a sum of two electric fields: the initial one, and the one from the conductor. So, you ask yourself, what does the electric field look like on the outside of the conductor when I know it has this "opposite field" on the inside. If you add both outside electric fields together, you see that it then results in that "reflection" behavior.

If I remember correctly, when you do the math, you also accidentally deduce Snell's Law.

 

[Noaha]

FYI, there is another way of looking at this, which is the superposition principle (a very useful principle overall if I might add).
So, you know that the electric field inside the conductor is zero. The way to look at it is by rather saying "the electric field *does* permeate the conductor, but additionally, the conductor creates a field inside that is the exact opposite."
The net result (of course) is still that the field inside is zero, but now you can look at the field *outside* the conductor as a sum of two electric fields: the initial one, and the one from the conductor. So, you ask yourself, what does the electric field look like on the outside of the conductor when I know it has this "opposite field" on the inside. If you add both outside electric fields together, you see that it then results in that "reflection" behavior.

If I remember correctly, when you do the math, you also accidentally deduce Snell's Law.

sorry for the late reply. I read your answer just now and I am not able to figure out what you are trying to tell? I m able to relate the situation with snell law(mathematically) but not able to physically see it.
Can you please further help out?

 

[Noaha]

FYI, there is another way of looking at this, which is the superposition principle (a very useful principle overall if I might add).
So, you know that the electric field inside the conductor is zero. The way to look at it is by rather saying "the electric field *does* permeate the conductor, but additionally, the conductor creates a field inside that is the exact opposite."
The net result (of course) is still that the field inside is zero, but now you can look at the field *outside* the conductor as a sum of two electric fields: the initial one, and the one from the conductor. So, you ask yourself, what does the electric field look like on the outside of the conductor when I know it has this "opposite field" on the inside. If you add both outside electric fields together, you see that it then results in that "reflection" behavior.

If I remember correctly, when you do the math, you also accidentally deduce Snell's Law.

I did some digging into this and found something about waves in ropes. when we produce wave in rope which is fixed from one end, the amplitude of the wave at the fixed point has to be zero all the time. So when a wave propagates and reaches the fixed point with some amplitude, it tries to pull the fixed point in the direction of its propagation. But the amplitude of fixed point being always zero, it inverts the propagation of wave and so the wave reflects with an inversion.

Are you trying to relate this rope situation with the EM waves hitting the conductor surface

 

[Noaha]

If you just think about a simple mirror. At the surface, the incident and reflected rays are 180 degrees out of phase, so they cancel. At a small height above the surface, the incident ray is slightly phase advanced, and the reflected ray is slightly phase delayed, so they no longer cancel.
At a height of a quarter of a wavelength, the incident ray is advanced in phase by 90 degrees and the reflected ray is delayed in phase by 90 degrees. So the time-of-travel has inserted 180 degree phase shift. Add to this the 180 degrees at the reflection point and you find that they are 360 degrees out of phase. In other words, in phase. And the two electric fields add vectorially, giving twice the field strength.

I will do a little more digging into your answer and ask again him I have any doubt.

 

[Vanadium 50]

Sorry about the delay.

If I have radiation, somewhere I have an oscillating charge. If you want to analyze the system by Method of Images, there is an image charge "inside" the conductor, and it must be oscillating as well, with the same frequency. At the exact moment the radiation from the real charge enters the conductor, the radiation from the image charge exits the conductor. Voila! Reflection! And of course, all boundary conditions are automatically satisfied.

 

[Noaha]

Sorry about the delay.

If I have radiation, somewhere I have an oscillating charge. If you want to analyze the system by Method of Images, there is an image charge "inside" the conductor, and it must be oscillating as well, with the same frequency. At the exact moment the radiation from the real charge enters the conductor, the radiation from the image charge exits the conductor. Voila! Reflection! And of course, all boundary conditions are automatically satisfied.

Nice explanation. I was not knowing that method of images can be applied on oscillating charges too. I will research more about method of image and get back to you if I have doubt. Thanks.